# Linear Differential Equation using Reduction of Order

I have this differential equation : $$4x^{2}y''+8x^{3}y'+(4x^{2}-3)y = 0$$ and it is given that the equation has a solution $$y_{1}(x)=x^{r}.$$

I set $y_2$ as $$y_{2}=x^{r}v(x).$$

When I'm trying to find the first and second derivative I start to fail.

I get $$y'_2=x^{r-1}(xv'+rv)$$ but the second is always long and makes no sense when I keep on with the problem. After I have put derivatives for the $y$, $y'$ and $y''$ in the original equation I get complex equations where i see no way to use reduction of order.

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Let's assume that we have the following equation: $y''+a(x)y'+b(x)y=0$.

If we know that $y_1(x)$ is a solution and we search our general solution in the form of $y_2(x)=y_1(x)v(x)$, then we get the following:

$y'_2=y'_1v+y_1v'$ and $y''_2=y''_1v+y'_1v'+y'_1v'+y_1v''=y_1''v+2y'_1v'+y_1v''$

Then by putting those values in the equation, we get

$y''_1v+2y'_1v'+y_1v''+a(x)(y'_1v+y_1v')+b(x)y_1v=0$

or

$(y''_1+a(x)y'_1+b(x))v+(2y'_1+a(x)y_1)v'+y_1v''=0$

As we know that $y_1$ is a solution therefore we have that

$y''_1+a(x)y'_1+b(x)=0$

So from our equation we get

$(2y'_1+a(x)y_1)v'+y_1v''=0$

$v''+(2y'_1/y_1+a(x))v'=0$

Now, in our example by putting $y=x^r$ in the equation we get

$4r(r-1)x^r+8rx^{r+2}+4x^{r+2}-3x^r=0$ from where we get $r=-1/2$.

From here it follows that solving our equation is the same as solving the following equation:

$y''+2xy'+y-3y/{4x^2}=0 => a(x)=2x$. Thus by taking $y=x^rv$ we get

$v''+(2(x^{-1/2})'/(x^{-1/2})+2x)v'=0 =>$

$v''+(x^{-1}+2x)v'=0=>$

Take $w=v'$ and get $w'+(x^{-1}+2x)w=0$ we are done.

I think the continuation is simple.

Sincerely,

Tigran

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It's probably easier if you first determine the (unique) value of $r$ which makes $y_1$ a solution.

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